2182 Lost Cows 已知逆序数求原来数列序列树状数组

最新推荐文章于 2021-02-19 21:55:20 发布

kongming_acm

最新推荐文章于 2021-02-19 21:55:20 发布

阅读量704

点赞数

CC 4.0 BY-SA版权

分类专栏： acm_数据结构文章标签： search output input each integer up

本文链接：https://blog.youkuaiyun.com/kongming_acm/article/details/5809683

acm_数据结构专栏收录该内容

112 篇文章

订阅专栏

本文介绍了一种解决特定逆序数序列问题的方法，通过树状数组和二分查找技术来确定一组牛的正确排序。该算法适用于竞赛编程中涉及逆序数计算的问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Lost Cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5119		Accepted: 3210

Description

N (2 <= N <= 8,000) cows have unique brands in the range 1..N. In a spectacular display of poor judgment, they visited the neighborhood 'watering hole' and drank a few too many beers before dinner. When it was time to line up for their evening meal, they did not line up in the required ascending numerical order of their brands.

Regrettably, FJ does not have a way to sort them. Furthermore, he's not very good at observing problems. Instead of writing down each cow's brand, he determined a rather silly statistic: For each cow in line, he knows the number of cows that precede that cow in line that do, in fact, have smaller brands than that cow.

Given this data, tell FJ the exact ordering of the cows.

Input

* Line 1: A single integer, N

* Lines 2..N: These N-1 lines describe the number of cows that precede a given cow in line and have brands smaller than that cow. Of course, no cows precede the first cow in line, so she is not listed. Line 2 of the input describes the number of preceding cows whose brands are smaller than the cow in slot #2; line 3 describes the number of preceding cows whose brands are smaller than the cow in slot #3; and so on.

Output

* Lines 1..N: Each of the N lines of output tells the brand of a cow in line. Line #1 of the output tells the brand of the first cow in line; line 2 tells the brand of the second cow; and so on.

Sample Input

Sample Output

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int n;//树状数组大小 
int c[8005];//树状数组 
int a[8005];//逆序数 
int lowbit(int x)
{
    return x&(-x);
}
void update(int x,int val)
{
    for(int i=x;i<=n;i+=lowbit(i)) 
    {
        c[i]+=val;
    }
}   
int getsum(int x)
{
    int cnt=0;
    for(int i=x;i>=1;i-=lowbit(i))
    {
        cnt+=c[i];
    }
    return cnt;
}   
int binary_search(int l,int r,int goal)
{
    if(l<r)
    {
        int mid=(l+r)/2;
        int cnt=getsum(mid);
        if(cnt==goal) return mid;
        else if(cnt>goal) return binary_search(l,mid,goal);
        else return binary_search(mid+1,r,goal);
    }
    else return -1;  
}      
int main()
{
    while(scanf("%d",&n)==1)
    {
        update(1,1);
        a[1]=0;
        for(int i=2;i<=n;i++) //这道题是从第2个开始 
        {
            scanf("%d",&a[i]);//输入逆序数 
            update(i,1);//更新树状数组 
        }
        for(int i=n;i>=1;i--)
        {
            int j=binary_search(1,n+1,a[i]+1);//二分查找 
            while(j>=1&&getsum(j)==a[i]+1) j--;j++;  
            //important   当出现1 2 3 3 3 4这种情况时候  应该找第一个3 
            a[i]=j;
            update(j,-1);//删除节点 
        }
        for(int i=1;i<=n;i++) printf("%d/n",a[i]);
    }
    return 0;
}